If $2m,\ m^{2}-1,\ m^{2}+1$ is a Pythagorean triple and $m = 4$, the largest member is

If $2m,\ m^{2}-1,\ m^{2}+1$ is a Pythagorean triple and $m = 4$, the largest member is
  1. $17$
  2. $15$
  3. $8$
  4. $9$

Solution

$m^{2} + 1 = 16 + 1 = 17$ is the largest. (Triple is $(8, 15, 17)$.)

Asked in: IMO

Practice more SQUARES AND SQUARE ROOTS questions on Aicharya